Problem: Let $h(x)=\sqrt[3]{x^2}$. $h'(x)=$
Solution: The strategy We can first rewrite $h(x)$ as a rational power of $x$. Then, the derivative of $h$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Rewriting the radical as a rational power $\begin{aligned} h(x)&=\sqrt[3]{x^2} \\\\ &=x^{^{\frac{2}{3}}} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}h'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{2}{3}}}\right) \\\\ &=\dfrac{2}{3}x^{^{\frac{2}{3}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac{2}{3}x^{^{-\frac{1}{3}}} \end{aligned}$ In conclusion, we found that $h'(x)=\dfrac{2}{3}x^{^{-\frac{1}{3}}}$. This can also be written as $\dfrac{2}{3\sqrt[3] x}$ (all equivalent forms are accepted).